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Let $f:(a,\infty)\rightarrow \mathbb{R}$ be a real-analytic and strictly monotone function. I have been wondering how much this function can "oscillate". Namely, can we always find a constant $C>1$ such that $f(2x)\leq C f(x)$ for all $x\in (a,\infty)$?

Of course, one obvious obstruction is fast growth (say $f(x)=e^x)$. However, what happens in the case if $f(x)\leq cx^n$ for some $c,n>0$? In this case the function cannot simply explode, but would need to have "large plateaus" where the function does not grow too much and then suddently it should grow again a lot. Intuitively, this should be prevented by the real-analyticity of the function, but I do not quite know how to approach this question. Are there tools (respectively areas of mathematics) that might help tackeling these type of questions?

As it is not quite clear which tools might be helpful, I'd appreciate anybody adding suitable tags.

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  • $\begingroup$ I seem to remember seeing a theorem saying that for any two continuous functions $g_1 < g_2$ there exists a real-analytic function $g$ with $g_1 < g < g_2$, then we can put $f = \int g$, if $g_1$ is positive it will be monotone and $\int g_1 \le \int g \le \int g_2$, and for these integrals it is not too hard to cook up the explosion you need. $\endgroup$
    – Aleksei Kulikov
    Commented Oct 19 at 23:49
  • $\begingroup$ @AlekseiKulikov Thanks, you seem to refer to something like this math.stackexchange.com/questions/2561363/…. I'll check the Carleman paper, this seems to be an incredibly strong statement. $\endgroup$
    – Severin Schraven
    Commented Oct 20 at 0:11
  • $\begingroup$ Yes, this one. It seems to answer your question (and any question of this sort) in the negative, but I'm too lazy to work out the details. $\endgroup$
    – Aleksei Kulikov
    Commented Oct 20 at 0:22
  • $\begingroup$ @AlekseiKulikov Absolutely, it just goes against all what I believed about real-analytic function. They are nothing like their complex-analytic siblings :( $\endgroup$
    – Severin Schraven
    Commented Oct 20 at 0:24
  • $\begingroup$ I'm not sure so don't quote me on that, but I think Carleman theorem is true even for entire functions? That is, you can even construct such an entire function. To get some positive results you need to impose some growth restriction on the whole of $\mathbb{C}$, then you can get something. $\endgroup$
    – Aleksei Kulikov
    Commented Oct 20 at 1:07

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An elementary example is given by the formula $$f(x)=\sum_{k=1}^\infty k!\,\Phi(x-k!)$$ for real $x>1$, where $\Phi$ is the standard normal cumulative distribution function. Then the function $f$ is real analytic on $(1,\infty)$ and $f(x)=O(x)$ for real $x>1$, whereas for natural $m$ we have $$\frac{f(m!)}{f(m!/2)}\sim\frac{m!/2}{(m-1)!}=\frac m2\to\infty$$ as $m\to\infty$.

The function $f$ can even be extended to the entire function $g$ defined by the formula $$g(z):=f(0)+\int_0^z dw \,\sum_{k=1}^\infty\frac{k!}{\sqrt{2\pi}}\,e^{-(w-k!)^2/2}$$ for complex $z$.

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  • $\begingroup$ Thank you, that's very instructive. $\endgroup$
    – Severin Schraven
    Commented Oct 20 at 11:45

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